166
6 Two-Dimensional Problems in Elasticity …
and
total shearing strain, γ r θ =
∂v
∂r
−
v
r
+
1
r
∂u
∂θ
(6.9c)
Differentiating Eq. (6.9a) with respect to θ and Eq. (6.9b) with respect to r, we
get
∂ε r
∂θ
=
∂
2 u
∂r ∂θ
(6.9d)
∂ε θ
∂r
=
1
r
∂u
∂r
−
1
r 2
u +
1
r
.
∂
2
v
∂r ∂θ
−
1
r 2
.
∂v
∂θ
=
ε r
r
+
1
r
∂
2
v
∂r ∂θ
−
1
r
u
r
+
1
r
∂v
∂θ
∴
∂ε θ
∂r
=
ε r
r
+
1
r
.
∂
2
v
∂r ∂θ
−
1
r
ε θ
(6.9e)
Now, differentiating Eq. (6.9c) with respect to r and using Eq. (6.9d), we get
∂γ r θ
∂r
=
∂
2
v
∂r 2 −
1
r
∂v
∂r
+
v
r 2 +
1
r
∂
2 u
∂r ∂θ
−
1
r 2
∂u
∂θ
=
∂
2
v
∂r 2 −
1
r
∂v
∂r
−
v
r
+
1
r
∂u
∂θ
+
1
r
∂
2 u
∂r ∂θ
∴
∂γ r θ
∂r
=
∂
2
v
∂r 2 −
1
r
γ r θ +
1
r
∂ε r
∂θ
(6.9f)
Differentiating Eq. (6.9e) with respect to r and Eq. (6.9f) with respect to θ , we
get
∂
2
ε θ
∂r 2 =
1
r
∂ε r
∂r
−
1
r 2
ε r +
1
r
∂
3
v
∂r 2 ∂θ
−
1
r 2
∂
2
v
∂r ∂θ
− −
1
r
∂ε θ
∂r
+
1
r 2 ε θ
(6.9g)
and
∂
2
γ r θ
∂r ∂θ
=
∂
3
v
∂r 2 ∂θ
−
1
r
∂γ r θ
∂θ
+
1
r
∂
2
ε r
∂θ 2
or
1
r
∂
2
γ r θ
∂r ∂θ
=
1
r
∂
3
v
∂r 2 ∂θ
−
1
r 2
∂γ r θ
∂θ
+
1
r 2
∂
2
ε r
∂θ 2
(6.9h)